Answer:
1. 
2. ![x=\pm7i[tex]3.[tex]\pm 12x =a](https://tex.z-dn.net/?f=x%3D%5Cpm7i%5Btex%5D%3C%2Fp%3E%3Cp%3E3.%5Btex%5D%5Cpm%2012x%20%3Da)
4. 22
Step-by-step explanation:
1. 



2. 


![x=\pm7i[tex]3. Area of square = [tex]a^{2}](https://tex.z-dn.net/?f=x%3D%5Cpm7i%5Btex%5D%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E3.%20Area%20of%20square%20%3D%20%5Btex%5Da%5E%7B2%7D)
Where a is the side
We are given an area o square = 
So, 


4. We are given that the solution of quadratic equation -9 and 9
So, equation becomes:


So, in the given equation 
z should be the number from which if we subtract 103 so we get 81
Substitute z = 22



Thus the value of z is 22
Answer:
we can not reject any value
Step-by-step explanation: From data we can test the highest and the lowest value to evaluate if one of these values are out of certain confidence Interval
If we established CI = 95 % then α = 5 % and α/2 = 0,025
From data we find the mean of the values
μ₀ = 12,03 and σ = 0,07
From z table we find z score for 0,025 is z(c) = ± 1,96
So limits of our CI are:
12,03 + 1,96 = 13,99
12,03 - 1,96 = 10,07
And all our values are within ( 10,07 , 13,99)
So we can not reject any value
Https://mathbitsnotebook.com/Algebra1/RealNumbers/RNProp.html